Zero function from an inequality

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $|f'(x)| \leq |f(x)|, \;\; \forall x \in \mathbb{R}$ and $f(0)=0$. Show that $f$ is the zero function.

Solution



We give three solutions

1st solution:

Successively we have that:

$$\begin{aligned}\left ( e^{-2x}f^2 (x) \right )' &=2e^{-2x}\left ( f(x)f'(x)-f^2(x) \right ) \\
&\leq 2e^{-2x}\left ( \left | f(x)f'(x) \right |-f^2(x) \right ) \\ &=2e^{-2x}\left | f(x) \right |\left ( \left | f'(x)\right | -\left | f(x) \right |\right ) \leq 0 \end{aligned}$$

Hence $e^{-2x} f^2(x)$ is decreasing and we get the result. 

2nd solution:

It is sufficient to prove the result on an arbitrary interval $[a, b]$. Let $M$ be the maximum of $|f|$ on the given interval. Therefore:

$$\begin{aligned}\left | f(x) \right | &= \left | \int_{0}^{x}f'(t)\, {\rm d}t \right |\\&\leq \int_{0}^{x}\left | f'(t) \right |\, {\rm d}t \\ &\leq \int_{0}^{x}|f(x)| \, {\rm d}x \\&\leq \int_{0}^{x}M \, {\rm d}x= Mx\end{aligned}$$

Inductively we show that (by repeating the procedure replacing $M$ with $Mx$):

$$\left | f(x) \right |\leq \frac{M x^n}{n!}\overset{x=A}{\Longrightarrow}\left | f(x) \right |\leq \frac{M A^n}{n!}\rightarrow 0$$

proving the result for $x\geq 0$. For $x<0$ we work with $f(-x)$ instead. 

 3rd solution:

We are invoking the basic inequality:

$$\left | \int_{a}^{b}g(t)\, {\rm d}t \right |\leq \int_{a}^{b}\left | g(t) \right |\, {\rm d}t  \tag{1}$$

Then successively we have:

 $$\begin{aligned}\left ( e^{-x}\int_{0}^{x}\left | f(t) \right |\, {\rm d}t \right )' &=e^{-x}\left ( \left | f(x) \right |- \int_{0}^{x}\left | f(t) \right |\, {\rm d}t \right ) \\&\overset{(1)}{\leq} e^{-x}\left ( \left | f(x) \right |- \int_{0}^{x}\left | f'(t) \right |\,{\rm d}t \right ) \\ &= e^{-x}\left ( \left | \int_{0}^{x}f'(t)\, {\rm d}t \right| - \int_{0}^{x} \left | f'(t) \right |\, {\rm d}t\right )\\&\leq 0\end{aligned}$$

meaning that the function on the left is decreasing, hence:

 $$e^{-x}\int_{0}^{x}\left | f(t) \right |\, {\rm d}t \leq \int_{0}^{0}\left | f(t) \right |\, {\rm d}t =0 , \;\;\; \forall x \geq 0$$

and the result follows for $x\geq 0$. For $x<0$  we work with $f(-x)$ instead.

The exercise can be found also in mathematica.gr